快速幂算法及矩阵快速幂

Description
给定三个数A, B, K, 求 A的B次方除以K的余数 。
Input
输入只有一行，为三个正整数A(1 <= A <= 2000000000), B(1 <= B <= 2000000000), K(1 <= K <= 10000)。
Output
一个整数，(A ^ B) % K 的值。
Sample Input
11 100 7
Sample Output

4

#include<iostream>using namespace std;int main(){long long a,b;int k,i,j,s=1;cin>>a;cin>>b;cin>>k;a=a%k;while(b){if(b%2==1)s=(s*a)%k;b=b/2;a=(a*a)%k; }cout<<s<<endl;return 0;}

以下为a^b快速幂算法模板

int s=1,a,b;        cin>>a>>b;while(b){if(b%2==1)<span style="white-space:pre"></span>s=s*a;b=b/2;a=a*a; }cout<<s<<endl;

使用矩阵快速幂算法求斐波那契数列

Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. F

Input

a single line containing n (where 0 ≤ n ≤ 100,000,000,000)

Output

print Fn mod 1000000007 in a single line.

Sample Input

99999999999

Sample Output

669753982

Hint

An alternative formula for the Fibonacci sequence is


As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by


Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

AC代码//矩阵快速幂模模板

#include<iostream>#include<cstring>#include<cstdlib>using namespace std;long long a[3][3]={{0,0,0},{0,1,1},{0,1,0}};long long t[3][3]={{0,0,0},{0,0,0},{0,0,0}};//相当于temp，替换作用long long s[3][3]={{0,0,0},{0,1,0},{0,0,1}};//初始为单位矩阵int main(){int k,i,j;long long n,b;cin>>n;b=n;while(b){if(b%2==1){    memset(t,0,sizeof(t));for(i=1;i<=2;i++)for(j=1;j<=2;j++) for(k=1;k<=2;k++) t[i][j]=(t[i][j]+(s[i][k]*a[k][j])%1000000007)%1000000007;//t=s*afor(i=1;i<=2;i++)for(j=1;j<=2;j++)s[i][j]=t[i][j];//s=t;}b=b/2;memset(t,0,sizeof(t));for(i=1;i<=2;i++)for(j=1;j<=2;j++)for(k=1;k<=2;k++)t[i][j]=(t[i][j]+(a[i][k]*a[k][j])%1000000007)%1000000007;//t=a*a;for(i=1;i<=2;i++)for(j=1;j<=2;j++)a[i][j]=t[i][j];//a=t;}cout<<s[1][2]<<endl;return 0;}

Problem Description 
Now, have a break, I, Jason, am tired of making the problem. 
Let us solve a math problem. 
F[1] = 1; 
F[2] = 1; 
F[n] = (A * F[n - 1] + B * F[n - 2]) mod 1000000007 
Given A, B, and n, you are to calculate the value of F[n]. 
 
Input 
The input consists of multiple test cases. Each test case contains three 
integers A, B and n on a single line (1<=A,B<=1000,1<=n<=100000000). 
  Three zeros signal the end of input and this test case is not to be 
processed. 
 
Output 
For each test case, print the value of F[n] on a single line. 
 
Sample Input 

1 1 3 
1 2 10 
0 0 0 
 
Sample Output 
2 
341 

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<set>#include<algorithm>#include<sstream>#define LL long longusing namespace std;const int MAX=1000000007;LL a[2][2];LL t[2][2];int main(){    LL x;    while(cin>>x)    {        LL y,n;        cin>>y>>n;        if(x==0&&y==0&&n==0)            return 0;        LL b=n-2;        LL s[2][2]={{1,0},{0,1}};        a[0][0]=x;        a[0][1]=y;        a[1][0]=1;        a[1][1]=0;        while(b)        {            if(b%2==1)            {            memset(t,0,sizeof(t));            for(int i=0;i<=1;i++)                for(int j=0;j<=1;j++)                    for(int k=0;k<=1;k++)                        t[i][j]=(t[i][j]+s[i][k]*a[k][j])%MAX;            for(int i=0;i<=1;i++)                for(int j=0;j<=1;j++)                    s[i][j]=t[i][j];            }        b=b/2;        memset(t,0,sizeof(t));        for(int i=0;i<=1;i++)            for(int j=0;j<=1;j++)                for(int k=0;k<=1;k++)                    t[i][j]=(t[i][j]+a[i][k]*a[k][j])%MAX;        for(int i=0;i<=1;i++)            for(int j=0;j<=1;j++)                a[i][j]=t[i][j];        }        cout<<(s[0][0]+s[0][1])%MAX<<endl;    }    return 0;}